predict shift in equilibrium position and effect on amount of chlorine gas when volume is doubled at constant temperature.

PCl3 (g) + Cl2(g) -> PCl5 (g)

OK. You need to get Le Chatlier's Principle CLEARLY in mind. In plain, but not very fancy, words, it says that a system in equilibrium will move to the left or to the right so as to undo what we do to it. If volume is doubled that means the pressure is halved (meaning it is less that it was). You have 2 mols gas on the left and 1 mol on the right. An INCREASE in P means the reaction will move to the side with fewer mols; the opposite if you have a DECREASE in P. So moving to the left means (PCl5) gets smaller and (Cl2) and (PCl3) become larger.

Here is a much longer way to do it but it has numbers and you may get the idea better.

.......PCl3 + Cl2 ==> PCl5
Let me just make up some numbers. Let's say PCl5 was 20, PCl3 was 10 and Cl2 was 10 when the system was at equilibrium. Then what would Keq be? That is
K = (PCl5)/(PCl3)(Cl2). Substituting the values I made up we get a K value of 0.2. That is from (20/(10)(10) = 0.2.

Now if we double the volume that means the concentrations are halved which makes the K expression = (10)/(5)(5) and that is 0.4. But remember that K is a constant and can't change. That's why the reaction must shift so as to keep the K from changing (in this case keep it at 0.2. So we have a value of 0.4 which is too high. That means the numerator must get smaller and/or the denominator must get larger(or both) in order to make that 0.4 go back to 0.2. So PCl5 must get smaller; Cl2 and PCl3 must get larger and that means a shift to the left.

To predict the shift in equilibrium position and the effect on the amount of chlorine gas when the volume is doubled at constant temperature, we need to analyze the reaction and understand how changes in volume affect the equilibrium.

In the given equation: PCl3 (g) + Cl2(g) -> PCl5 (g)

When the volume is doubled, the reaction will experience a change in pressure. However, since the temperature is constant, the change in volume will not affect the total number of moles of gas present.

There are two possible scenarios based on Le Chatelier's principle, which states that a system at equilibrium will shift to counteract any changes made to it.

1. Increase in volume: Doubling the volume will decrease the pressure. As a result, the system will shift towards the side with more moles of gas to increase the pressure. In this case, the right side of the equation has two moles of gas (PCl5), while the left side has two moles (PCl3 + Cl2). Thus, the equilibrium will shift to the right, favoring the formation of more PCl5.

2. Decrease in volume: If the volume is halved, the pressure will increase, and the equilibrium will shift to the side with fewer moles of gas to decrease the pressure. In this case, the equilibrium will shift towards the left, favoring the formation of more PCl3 and Cl2, and reducing the amount of PCl5.

To summarize, when the volume is doubled at constant temperature in this specific reaction, the equilibrium will shift to the right, favoring the formation of more PCl5, while reducing the amounts of PCl3 and Cl2.